Calculating Displacement of Uniform Bar at Centre with Respect to Time

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SUMMARY

The discussion focuses on calculating the displacement "s" of the center of gravity of a uniform bar supported by two springs with differing stiffness. One spring has twice the stiffness of the other, leading to complex oscillatory behavior. The problem requires the application of normal modes derived from coupled differential equations to analyze the interaction between the springs and the bar. Special cases, such as equal springs and varying spring constants, are suggested for simplification and understanding of the system's dynamics.

PREREQUISITES
  • Understanding of simple harmonic motion
  • Knowledge of coupled differential equations
  • Familiarity with the concept of normal modes
  • Basic principles of spring mechanics and equivalent spring constants
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  • Study the derivation of normal modes in coupled oscillators
  • Learn about the calculation of equivalent spring constants for springs in series and parallel
  • Explore the concept of center of mass in oscillatory systems
  • Investigate the effects of varying spring stiffness on oscillation frequencies
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Numbers123
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A uniform bar of length "l" and mass "m" is supported at its ends by two springs. One spring has twice the stiffness of the other. From the equilibrium position the centre of the bar is pulled down a short distance and then released. How do you calculate the displacement "s" of the centre of gravity with respect to time?
This is a self inflicted question after looking at the vibrations of weights on springs. I am not a student (63 years old) just interested in physics.
I am assuming that the springs at either end of the bar will oscillate with simple harmonic motion of different frequencies (one twice the other) but how do they combine at the centre.
 
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That problem is a bit complicated. It as to be solved using "normal modes", which follow from coupled differential equations.
 
Numbers123 said:
A uniform bar of length "l" and mass "m" is supported at its ends by two springs. One spring has twice the stiffness of the other. From the equilibrium position the centre of the bar is pulled down a short distance and then released. How do you calculate the displacement "s" of the centre of gravity with respect to time?
This is a self inflicted question after looking at the vibrations of weights on springs. I am not a student (63 years old) just interested in physics.
I am assuming that the springs at either end of the bar will oscillate with simple harmonic motion of different frequencies (one twice the other) but how do they combine at the centre.

This is a very complicated problem you have self-inflicted. But it's interesting to think about.

First just think about some special cases. Consider the case where both springs are the same. What happens when springs are in parallel? What would be the equivalent spring constant? How is it different from springs in series? What would be the spring constant then? What happens if two springs are replaced by one spring at the center of the bar? How does the bar oscillate if it is pulled and released without tilting it? Can you use the concept of "center of mass" then to assume all the mass of the bar is at one point? What happens if you have two equal springs and you pull the bar equally and release it. What happens if you have two equal springs on the end and you lower one side and raise the other side and release it? Are the resonant frequencies the same in both cases? (springs and mass have not changed) Explain the result. What happens if the springs are not equal? Can the motion of each end of the bar be separated from that of the other end? Does it help if one assumes the vibrations of the bar are of very small amplitude. What happens to your forces as the amplitude of the bar motion gets very large (has to do with the angles of the forces a the point of spring attachments)?

Do you see what a can of worms this "simple" problem is? Getting the right answer is going to be all about asking the right questions.
 

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